Area of a Polar Region or the Area Bounded by a Single Polar Curve: Study Guide (Calculus BC 9.8)

BC-only material. Polar area is part of Calculus BC only. Calculus AB students do not need this topic.

Before you start: prerequisites

You need polar curves from Topic 9.7 and the idea of an integral as a limit of sums. If any row is shaky, revisit it from the Calculus BC roadmap first.

Prerequisite	Topic	What you use it for here
Polar coordinates and curves	9.7	Reading and sketching the region
Riemann sums and definite integrals	6.2–6.4	Turning a sum of sectors into an integral
Sector area ½r²θ (radians)	Geometry	The building block of the formula
Power-reducing identities	Precalculus	Integrating sin²(kθ) and cos²(kθ) by hand

Notation on this page: “½ ∫ from α to β of r² dθ” is a definite integral with lower limit α and upper limit β. Angles are in radians.

Why rectangles do not fit

For y = f(x), you found areas by slicing into thin vertical rectangles, each of area f(x) Δx. A polar region is different. Its natural boundaries are rays from the pole and a curve r = f(θ). Thin rectangles do not line up with rays. Thin sectors do, like slices of a pizza.

A sector of a circle. A sector with radius r and angle θ (in radians) is the fraction θ/(2π) of a full disc, so its area is

(θ / 2π) × πr² = ½ r² θ

For example, a sector with radius 4 and angle π/6 has area ½ × 16 × π/6 = 4π/3.

Building the formula from thin sectors

Take the region bounded by the curve r = f(θ) and the rays θ = α and θ = β, where f is continuous and the region is swept out once as θ goes from α to β.

1. Split [α, β] into n small angles, each Δθ wide.
2. On each small angle, the curve is close to a circular arc with radius r = f(θₖ) for some θₖ in that piece. So that slice has area about ½ f(θₖ)² Δθ.
3. Add the slices: area ≈ Σ ½ f(θₖ)² Δθ. This is a Riemann sum for the function ½ f(θ)².
4. Let n → ∞ (so Δθ → 0). The sum becomes a definite integral:

Area = ½ ∫ from α to β of r² dθ = ½ ∫ from α to β of [f(θ)]² dθ

This is the same idea you used for rectangular areas: approximate with simple shapes, add, take the limit. Only the shape of the slice has changed.

Area as an accumulation. If A(θ) is the area swept out from α up to angle θ, the Fundamental Theorem of Calculus gives dA/dθ = ½ r². So ½ r² is the rate at which area is being swept out, in square units per radian.

Choosing the limits: trace the region once

The integral is easy. The limits are where the marks are won or lost. Use three tools.

1. Zeros of r mark the ends of petals and loops. A petal or loop starts and ends at the pole, where r = 0. For the three-petal rose r = 2 cos 3θ, solve cos 3θ = 0: θ = ±π/6 are consecutive zeros, and r > 0 between them. So one petal is

½ ∫ from −π/6 to π/6 of 4 cos²3θ dθ = 2 ∫ from −π/6 to π/6 of ½(1 + cos 6θ) dθ = [θ + (1/6) sin 6θ] from −π/6 to π/6 = π/3.

All three petals together have area 3 × π/3 = π.

2. Check how many times the curve is traced. The rose r = 2 cos 3θ is traced completely for 0 ≤ θ ≤ π. Integrating from 0 to 2π gives 2π, which counts every petal twice. Likewise, the circle r = 6 cos θ is traced once for 0 ≤ θ ≤ π (area 9π, the area of a circle of radius 3) but twice for 0 ≤ θ ≤ 2π (giving 18π). A quick table of values, or the graph on a calculator, tells you when the curve starts repeating.

3. Use symmetry, but carefully. If the region is symmetric about the x-axis, you can integrate over the top half and double. Only do this when you are sure the two halves match; a sketch is the safest check.

Negative r. The formula squares r, so it measures swept area whatever the sign of r. But when r < 0, the points plotted are on the opposite side of the pole. Draw the region before choosing limits, so you measure the part you actually want.

Worked example 1: a limaçon by hand

Question. The curve r = 2 + cos θ is shown in Figure 1. Without a calculator, find

(a) the area of the whole region enclosed by the curve; (b) the area of the part of that region in the first quadrant.

(a) r = 2 + cos θ ≥ 1 > 0 for every θ, so the curve is a single loop traced once for 0 ≤ θ ≤ 2π.

1. Set up: Area = ½ ∫ from 0 to 2π of (2 + cos θ)² dθ.
2. Expand: (2 + cos θ)² = 4 + 4 cos θ + cos²θ.
3. Integrate each term over [0, 2π]:
   • ∫ 4 dθ = 8π
   • ∫ 4 cos θ dθ = 4[sin θ] = 0
   • ∫ cos²θ dθ = ∫ ½(1 + cos 2θ) dθ = [½θ + ¼ sin 2θ] = π
4. Area = ½(8π + 0 + π) = 9π/2 ≈ 14.137.

Check: the curve stays between r = 1 and r = 3, so the area must be between π(1)² ≈ 3.14 and π(3)² ≈ 28.27. 14.137 fits.

(b) The first quadrant is the region between the rays θ = 0 and θ = π/2.

1. Area = ½ ∫ from 0 to π/2 of (4 + 4 cos θ + cos²θ) dθ.
2. ∫ from 0 to π/2 of 4 dθ = 2π. ∫ from 0 to π/2 of 4 cos θ dθ = 4[sin θ] from 0 to π/2 = 4. ∫ from 0 to π/2 of cos²θ dθ = [½θ + ¼ sin 2θ] from 0 to π/2 = π/4.
3. Area = ½(2π + 4 + π/4) = 2 + 9π/8 ≈ 5.534.

Check: in the first quadrant 2 ≤ r ≤ 3, so the area lies between a quarter disc of radius 2 (π ≈ 3.14) and one of radius 3 (9π/4 ≈ 7.07). 5.534 fits, and it matches the six-sector estimate of about 5.540 in Figure 1.

Why is (b) more than a quarter of (a)? A quarter of 9π/2 is about 3.534. The curve bulges to the right (r is largest at θ = 0), so the right-hand half holds more of the area.

Worked example 2: a lawn sprinkler, with a calculator

Context. A fictional rotating lawn sprinkler sits at the pole. As it turns from θ = 0 to θ = π, it throws water a distance r = 4 + cos 3θ metres. The watered region is bounded by this curve and the x-axis.

(a) Find the watered area. r ranges from 3 to 5, so r > 0 and the region is swept out once on [0, π]. Area = ½ ∫ from 0 to π of (4 + cos 3θ)² dθ = ½ ∫ from 0 to π of (16 + 8 cos 3θ + cos²3θ) dθ. By hand: ∫ 16 dθ = 16π; ∫ 8 cos 3θ dθ = (8/3)[sin 3θ] from 0 to π = 0; ∫ cos²3θ dθ = π/2. Area = ½(16π + π/2) = 33π/4 ≈ 25.918 m². A calculator’s numerical integral agrees.

Check: the area is between half-discs of radius 3 (about 14.137) and 5 (about 39.270). ✓

(b) At what rate is area being watered, per radian of turning, when θ = π/3? dA/dθ = ½ r². At θ = π/3, r = 4 + cos π = 3, so dA/dθ = ½ × 9 = 4.5 m² per radian.

(c) The sprinkler is switched off once it has watered 10 m². At what angle does this happen? Solve ½ ∫ from 0 to k of (4 + cos 3θ)² dθ = 10 on a calculator (for example, graph the integral as a function of k, or use a solver). This gives k ≈ 1.328 radians. Sense check: at the start r = 5, so area builds at ½ × 25 = 12.5 m² per radian; slower later as r falls towards 3. Reaching 10 m² in a little over 1.3 radians is reasonable.

Write-up tip. On a calculator question, write the integral with its limits before giving the number. The setup is a mark on its own.

Common misconceptions

• “Area = ∫ r dθ.” You must square r and multiply by ½. ∫ r dθ is not an area of anything useful here.
• Forgetting the ½. The sector area is ½ r² θ, not r² θ.
• Squaring incorrectly. (2 + cos θ)² is 4 + 4 cos θ + cos²θ, not 4 + cos²θ.
• Integrating 0 to 2π by habit. Many curves repeat. Find the interval that traces the region once.
• Using the wrong zeros of r for a petal. Use consecutive zeros with the petal between them, and check the sign of r in between.
• Ignoring where negative r puts the points. The formula still gives a positive area, but possibly of a part of the curve you did not intend.
• Integrating sin²θ as −cos³θ/3 or similar. Use the power-reducing identity first.
• Degree mode on the calculator. Every formula here assumes radians.

Where this leads

Next, Topic 9.9 finds the area of a region between two polar curves: the area inside one curve and outside another is ½ ∫ (R² − r²) dθ, where R is the outer curve and the limits come from where the curves meet. Continue with Topic 9.9, Finding the Area of the Region Bounded by Two Polar Curves, or see the order on the Calculus BC roadmap.

Try the practice questions now, then use the revision notes and the checklist to consolidate.